Question

The rate constant for a certain reaction is measured at two different temperatures:

Temperature K
376.0°C 4.8 x 10^8
280°C 2.3 x 10^8

Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy for this reaction. Round your answer to significant digits.

Answer 1

Answer: 22.78KJ/ mol

Step-by-step explanation:

The computation of activation energy
`E_0` follows by formula:

`\ln ((K_2)/(K_1)) = ((E_a )/( R)) ((1)/(T_1) -( 1)/(T_2))`

Given:
`T_1 = 376.0^(\circ)C =376+273K= 649 K`

`K_1 = 4.8 *10^8`

`T_2 = 280.0^(\circ)C = 273+280K=553 K ,\ \ \ K_2 = 2.3 * 10^8`

`\ln((2.3 *10^8 )/( 4.8 * 10^8)) = ((E_a )/( 8.314)) ((1)/(649) - (1)/(553))\\\\\Rightarrow\ \ln(0.479) = ((E_a )/( 8.314)) (-(96)/(358897))\\\\\Rightarrow\ -0.73605468= ((E_a )/( 8.314)) (-(96)/(358897))\\\\\Rightarrow\E_a=-0.73605468* 8.314*(-358897)/(96)=22878.033\\\\\Rightarrow\E_a=22878.033\ J/mol=22.78 KJ /mol`

Hence, the activation energy for this reaction =22.78KJ/ mol